sols18

sols18 - 44 18. November 8th: Dedekind cuts. 18.1. Prove...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
44 18. November 8th: Dedekind cuts. 18.1 . Prove that the relation on D defned in Defnition 18.2 is antisymmetric. Answer. Assume ( A 1 | B 1 ) ( A 2 | B 2 ) and ( A 2 | B 2 ) ( A 1 | B 1 ). By defnition, this means A 1 A 2 and A 2 A 1 . Since is antisymmetric, this implies A 1 = A 2 . It then Follows that B 2 = A 2 = A 1 = B 1 , proving that ( A 1 | B 1 ) = ( A 2 | B 2 ). ± 18.2 . With notation as in the prooF oF Theorem 18.4, prove that x<y For all x L and all y M . Answer. Since x L (and by defnition oF union over Families, Defnition 5.7) there exists an index j I such that x A j . Since y M , y B i For all i I (by defnition oF intersection over Families), and in particular y B j . Since ( A j | B j ) is a Dedekind cut, it Follows that x < y , as required. ± 18.3 . Provide a sensible defnition oF the ‘sum’ ( A 1 | B 1 )+( A 2 | B 2 ) oF two arbitrary Dedekind cuts ( A 1 | B 1 ), ( A 2 | B 2 ). This should be done in such a way that iF q 1 ,q 2 Q and ι ( q 1 ) = ( A 1 | B 1 ), ι ( q 2 )=( A 2 | B 2 ), then ( A 1 | B 1 A 2 | B 2 )= ι ( q 1 + q 2 ).
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 12/14/2011 for the course MGF 3301 taught by Professor Aluffi during the Fall '11 term at FSU.

Ask a homework question - tutors are online